Download PHYS 342: Modern Physics

Example Problem #1.38
• For components of C, we have:
Cx = (3.1 km)(cos45o) = 2.2 km
Cy
45o
Cx
Cy = (3.1 km)(sin45o) = 2.2 km
• Components of total displacement R are thus:
R
q
Rx
Ry
Rx = Ax + Bx + Cx = 0 + 4.0 km + 2.2 km = 6.2 km
Ry = Ay + By + Cy = 2.6 km + 0 km + 2.2 km = 4.8 km
• Magnitude of R is:
|R| = (Rx2 + Ry2)1/2 = [(6.2 km)2 + (4.8 km)2]1/2 = 7.8 km
• Direction of R found from trig. functions:
sinq = 4.8 km/7.8 km  q = sin–1(4.8/7.8) = 38o
Example Problem #2.33
(a) max. speed attained when acceleration = 0  constant speed
achieved (aqua region)
 v = v0 + at  look at times when t = 0 (v0 = 0) and t = t1 (v = max.
and const.)
 v = at1  a = 20.0 m/s2  t1 = 15 min = 900 s
 v = (20.0 m/s2)(900 s) = 18,000 m/s = 18.0 km/s
(b) 1st leg of journey (represented by red line in graphs):
x – x0 = ½(v0 + v)t  x0 = 0, v0 = 0
 x = x1 = vt/2  v = 18.0 km/s  t = 900 s  x1 = 8100 km
3rd leg of journey (yellow line):
x – x0 = ½(v0 + v)t  x3 – x2 = ½(v2 + v3)t  v2 = 18.0 km/s, v3 = 0,
t = 900 s  x3 – x2 = 8100 km  leg 1 + leg 3 = 16,200 km
Example Problem #2.33
(b continued) leg 2 distance = 384,000 km – 16,200 km = 367,800 km
 fraction of total distance = leg 2 dist. / total dist. =
367,800 km / 384,000 km = 0.958
(c) Find time ship was traveling at constant speed (leg 2, aqua region):
x – x0 = ½(v0 + v)t  x2 – x1 = ½(v1 + v2)t
 x1 = 8100 km, x2 = 384,000 km – 8100 km = 375,900 km, v1 =
18.0 km/s, v2 = 18.0 km/s
 t = 2(x2 – x1) / (v1 + v2) = 20,433 s = tleg2
 ttotal = tleg1 + tleg2 + tleg3 = 900 s + 20,433 s + 900 s = 22,233 s
= 370.56 min = 6.18 hr
+y
443 m
Free Fallin’
How long would it take for Tom Petty to
go “Free-Fallin’” from the top of the
Sears Tower?
 y = y0 + v0t – (1/2)gt2
 – 443 m = 0 + 0 – (1/2) gt2
 886 m / g = t2  t2 = 90.4 s2
 t =  9.5 s  t = 9.5 s (neg. value has no physical meaning)
How fast will he be moving just before he hits the ground?
 v = v0 – gt = 0 – gt = – (9.8 m/s2)(9.5 s) = – 93.1 m/s
(negative sign means downward direction)
Example Problem #3.12
y
v0x = v0.
x
Let x0 = 0
and y0 = 0
v0y = 0.
x = x0 + v0xt.
y = y0 + v0yt – 0.5gt2.
What must v0 be so that x = 1.75 m when y = –9.00 m?
 Time it takes to travel 9.00 m vertically:
y = – 0.5gt2 = – 0.5(9.8 m/s2)t2 = –9.00 m  t = 1.36 s
 Speed to travel 1.75 m horizontally:
x = v0t  v0 = x / t = 1.75 m / 1.36 s = 1.29 m/s
Example Problem #3.34
(a) arad = v2 / R = (3 m/s)2 / (14.0 m) =
0.643 m/s2
atan = 0.5 m/s2
arad
atan = 0.5 m/s2
 a = [(arad)2 + (atan)2]1/2 = [(0.643 m/s2)2 +
(0.5 m/s2)2]1/2 = 0.814 m/s2
 Direction of a determined by:
tan q = arad / atan  q = tan–1 (0.643/0.5)  q = 52.10 (up
from horizontal)
(b)
a
52.10
v
Example Problem #4.39
(a) Since the two crates are connected by the light rope, they
move together with the same acceleration of 2.50 m/s2.
m = 4.00 kg
(b) y
N2: S Fx = T = max =
(4.0
2) = 10.0 N
T
kg)(2.50
m/s
x
mg
(c)
Nm
M = 6.00 kg
T
F
y
Mg
x
NM
Net force points in the +x –
direction (same direction as
acceleration), making force F
larger in magnitude
Example Problem #4.39 (continued)
(d)
M = 6.00 kg
T
F
y
Mg
NM
x
N2: S Fx = F – T = Max
F = Max + T = (6.00 kg)(2.50 m/s2) + 10.0 N = 25.0 N
Example Problem #5.7
Free–body diagram of wrecking ball:
TB
y
400 TBy
TA
mg
TBx
x
mg = (4090 kg)(9.8 m/s2)
= 40000 N
(a) S Fy = 0  TBy – mg = 0  TBcos400 – mg = 0
 TB = mg / cos400 = 5.23  104 N
(b) S Fx = 0  TBx – TA = 0  TBsin400 – TA = 0
 TA = TBsin400 = 3.36  104 N
Example Problem #5.10
v = constant
Piano moving at const. velocity
 a = 0 (piano in equilibrium)
110
Free–body diagram:
N
y
F
x
110
W
In general:
N
y
x
110 Wy
W
Wx
q
q
900 - q q
S Fx = 0  F – Wx = 0  F = Wx
 F = mg sin110 = (180 kg)(9.8 m/s2) sin110 = 336.6 N
VI.B. Dynamics Problems
1. Dynamics problems involve bodies which have
a nonzero acceleration


nd
2. From Newton’s 2 Law:  F  ma
3. In component form:  Fx  max  Fy  may
4. Summary of Problem–Solving Strategy:
a) Similar to strategy given for statics problems
(bodies in equilibrium)
b) Exception: In Step #3, set  Fx  max  Fy  may
as appropriate
Problem #5.16
Free–body diagram:
y
N
x
q
Apply Newton’s 2nd Law in x – direction:
Wy
q W
Wx
S Fx = max  Wx = max  Wsinq = max
 mg sinq = max  sinq = ax / g
From 1 – D motion with constant acceleration:
 v2 = v02 + 2a(x – x0)  a = (v2 – v02) / 2(x – x0)
Analysis of Swinging Pail of Water
Top:
vt
+y
Bottom:
vb
Free–body diagrams of water:
Fp
(top)
mg
Fp
(bottom)
mg
Force exerted on water by pail at top:
S Fy = may = m(–vt2 / r)  – Fp – mg = m(–vt2 / r) 
Fp = m (vt2 / r) – mg
Analysis of Swinging Pail of Water
Minimum value of vt for water to remain in pail:
 Minimum force pail can exert is zero, so set Fp = 0 and
solve for minimum speed vt,min:
0 - mg = m(–vt,min2 / r)  vt,min2 = rg  vt,min = (rg)1/2
Force exerted on water by pail at bottom:
S Fy = may = m(vb2 / r)  Fp – mg = m(vb2 / r) 
Fp = m(vb2 / r) + mg
Remember that centripetal force is not an external force
acting on a body – it is just the name of the net force acting
on a body undergoing circular motion (so there is no arrow
for centripetal force on a free–body diagram)
Example Problem #5.97
Free–body diagram of car:
N
W
y
R
f
x
(a) S Fx = max = m(v2/R)  f = m(v2/R)  R = mv2/f
 f = msN = msW = msmg  R = mv2/msmg = v2/msg =
(35.8 m/s)2/(0.76)(9.8 m/s2) = 171.7 m (about 563 ft.)
(b) From above, vmax2 = msgR  vmax = (msgR)1/2 =
[(0.20)(9.8 m/s2)(171.7 m)]1/2 = 18.34 m/s = 41.0 mph
(c) vmax = (msgR)1/2 = [(0.37)(9.8 m/s2)(171.7 m)]1/2 = 24.95
m/s = 55.8 mph
The posted speed limit is evidently designed for wet road
conditions.
Example Problem #6.23
(a) Free–body diagram of 12–pack :

s
y
N
F = 36 N
mg
x
Work–Energy Theorem: Wtot = K2 – K1 = ½ m(v22 – v12) = ½
mv22 (v1 = 0  starts from rest)
 Wtot = WN + Wmg + WF = WF
 WF = Fs cos00 = Fs = (36.0 N)(1.20 m) = 43.2 J = ½ mv22
 v2 = [2(43.2 J) / m]1/2 = [2(43.2 J) / 4.30 kg]1/2 = 4.48 m/s
(b) Free–body diagram of 12–pack :

s
N
y
F = 36 N
fk
mg
x
Example Problem #6.23 (continued)
 Wtot = Wf + WF
 Wf = fs cos1800 = –fs = – (mkN)s = –mkmgs
= –(0.30)(4.30 kg)(9.8 m/s2)(1.20 m) = –15.17 J
 Wtot = –15.17 J + 43.2 J = 28.03 J
 v2 = [2(28.03 J) / 4.30 kg]1/2 = 3.61 m/s
Example Problem
v0

s
N
v=0
f
mg
(a) Wtot = K2 – K1 = ½ m(v22 – v12) = –½ mv02 (v2 = 0)
 Wtot = Wf + WN + Wmg = Wf
 Wf = fs cos1800 = –fs = – (mkN)s = –mkmgs
 –mkmgs = –½ mv02  s = v02 / 2mkg
(b) For v0 = 80.0 km/h = 22.2 m/s:
mk = v02 / 2gs = (22.2 m/s)2 / 2(9.8 m/s2)(91.2 m) = 0.28
 For v0 = 60.0 km/h = 16.7 m/s:
s = (16.7 m/s)2 / 2(0.28)(9.8 m/s2) = 50.8 m
Example Problem #6.36
x = – 0.025 m
v2
x=0
(Spring compressed)
x=0
(Spring relaxed)
(a) W = 1/2 kx2 = 1/2 (200 N/m)(0.025 m)2 = 0.062 J
(b) Wtot = K2 – K1 = 1/2 m(v22 – v12) = 1/2 mv22 (block
initially at rest when spring is compressed)
 Wtot = WN + Wmg + WS = WS = 0.062 J
 0.062 J = 1/2 mv22  v2 = [2(0.062 J) / m]1/2
= [2(0.062 J) / (4.00 kg)]1/2 = 0.177 m/s
IX.B. Gravitational Potential Energy
1. Consider the fall of Tom Petty from the Sears Tower
again
a) Once Tom Petty begins to fall, gravity does work on him,
accelerating him toward the ground
b) Work done by gravity on Tom Petty:
 
Wg  Fg  s  Fg s cosq  Fg s  mgs  mgh (IX.B.1)
c) More generally:
Wg  mg  y1 - y2  (IX.B.2)
y

a
h (y1)
Free – body diagram
(neglect air resistance):

s
0 (y2)
W = Mg
IX.B. Gravitational Potential Energy
1. Product of weight mg and vertical height y is defined as
the gravitational potential energy U = mgy
a) Has units of (kg)(m/s2)(m) = J
2. Work done by gravity can thus be interpreted as a
change in the gravitational potential energy:
Wg  mg  y1 - y2   U1 - U 2  -U 2 - U1   -U
(II.B.3)
Path independent – all that matters is change in vertical position
Design of a Loop–the–Loop Roller Coaster
Suppose we wish to design the following Loop–the–Loop
1
roller coaster:
0
H – 2R
2
y
y1 = H
R
y2 = 2R
What is the minimum value of H such that the roller coaster
cars make it safely around the loop? (Assuming cars fall
under influence of gravity only.)
Conservation of mechanical energy:
1/2mv12 + mgy1 = 1/2mv22 + mgy2
Assume that roller coaster starts from rest at top of hill.
Then we have: mgH = 1/2mv22 + mg(2R)
v22 = 2mg(H – 2R) / m = 2g(H – 2R)
Design of a Loop–the–Loop Roller Coaster
For car to make it safely over the loop:
acar = arad  g (remember water in bucket)
v22 / R  g
2g(H – 2R) / R  g
H  R/2 + 2R, or
H  5R/2.
Example Problem
George
y1
0
y2
450 300
Conservation of mechanical energy:
1/2mv12 + mgy1 = 1/2mv22 + mgy2
y1 = –(20 m)cos450 = –14.14 m
y2 = –(20 m)cos300 = –17.32 m
v1 = 0 (George starts from rest)
+y
Example Problem #7.12 (continued)
So:
1/2 mv22 = mgy1 – mgy2
v22 = 2g(y1 – y2)
v2 = [2g(y1 – y2)]1/2
v2 = [2(9.8 m/s2)(–14.14 m – (–17.32 m))]1/2
v2 = 7.89 m/s
pt. 2
Example Problem #7.18
(a) Conservation of mechanical energy:
y2 = 22.0 m
pt. 1
y1 = 0
Ug = 0
K1 + U1,g + U1,el = K2 +U2,g + U2,el
 K1 = 0 (pebble initially at rest)
 U1,g = 0 (by choice)
 K2 = 0 (pebble at rest at max. height)
 U2,el = 0 (slingshot in relaxed position)
 U1,el = U2,g = mgy2 = (0.01 kg)(9.8 m/s2)(22.0 m) = 2.16 J
(b) U1,el = 2.16 J = U2,g = mgy
 y = 2.16 J / mg = 2.16 J / (0.025 kg)(9.8 m/s2) = 8.82 m
(c) No air resistance, no deformation of the rubber band
Example Problem #7.43
x1 = –0.20 m
x=0
Point 1 (Spring compressed) s = 1.0 m
Point 2 (Block stopped)
With friction doing work on the block, we have:
0
0
0
2
2
2
2
½ mv1 + ½ kx1 + Wf = ½ mv2 + ½ kx2
 Wf = – ½ kx12 = – ½ (100 N/m)(–0.20 m)2 = –2 J
Also, Wf = – f s = – f (1.00 m) = –2 J
 f = 2 N = mkN = mkmg
 mk = 2 N / mg = 2 N / (0.50 kg)(9.8 m/s2) = 0.41
Example Problem #7.73
(Spring compressed)
Pt. A
300
v = 7 m/s
Pt. B
L=6m
H
300
From the Work – Energy Theorem:
0
0
0
KA + UA,g + UA,el + Wf = KB + UB,g + UB,el  UA,el = KB + UB,g – Wf
 KB = ½ mvB2 = ½ (1.50 kg)(7.00 m/s)2 = 36.75 J
 UB,g = mgH = mgLsin(300) = (1.50 kg)(9.8 m/s2)(6.00 m)sin(300)
= 44.1 J
N
f
0
 Wf = – f s = – mkNL = – mkmgcos(30 )L
= – (0.50)(1.50 kg)(9.8 m/s2)cos(300)(6.00 m)
mg
= – 38.19 J
 UA,el = 36.75 J + 44.1 J – (–38.19 J) = 119.0 J
X. E. Collisions
1. Inelastic collisions
a) Classic example: car crash where cars stick together after
collision
b) vf?


Conservation of momentum: Pbefore  Pafter



Pbefore  m1v1i  0  m1v1i


Pafter  m1  m2 v f


m1v1i  m1  m2 v f

vf 
m1 
v1i
m1  m2
(X.E.1,2)
(X.E.3)
(X.E.4)
X. E. Collisions
a) Limiting behavior: m1 => 0: vf => 0
b) Limiting behavior: v1 => 0: vf => 0
c) Limiting behavior: m1 = m2: vf = 1/2 v1i
d) In inelastic collisions, the KE of system after collision
< KE of system before collision (Why?)
• K.E. before collision = K1 = 1/2 m1v1i2
• K.E. after collision = K2 = 1/2 (m1 + m2)vf2
= 1/2 (m1 + m2)[m1 / (m1 + m2)]2v1i2
• So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1
X. E. Collisions
2. Elastic collisions
a) Forces between colliding bodies are conservative
b) Kinetic energy is conserved (may be temporarily converted
to elastic potential energy)
c) Momentum is conserved
X. F. Elastic Collisions
1. Consider a head-on collision where one object is at
mB, vB1 = 0
rest at t=0:
mA, vA2
mA, vA1
mB, vB2
2. Conservation of kinetic energy gives:
1/2 mAvA12 = 1/2mAvA22 + 1/2mBvB22 (X.F.1)
3. Conservation of momentum gives:
mAvA1 = mAvA2 + mBvB2
(X.F.2)
4. Combining F.1 and F.2 yields:
v A2
m A - mB

v A1
m A  mB
vB 2
2m A

v A1 (X.F.3,4)
m A  mB
X. F. Elastic Collisions
5. Limiting Case: mA = mB
vA2 = 0 and vB2 = vA1
6. Limiting Case: mA << mB
vA2  – vA1
vB2 << vA1
Similar to result of ping-pong ball (mA) striking stationary
bowling ball (mB)
7. Limiting Case: mA >> mB
vA2  vA1
vB2  2vA1
Similar to result of bowling ball (mA) striking stationary
ping-pong ball (mB)
X. F. Elastic Collisions
8. “Grazing” collision:
mB, vB1 = 0
q
mA, vA1
mA, vA2
j
mB, vB2
In this case, we need to remember to take both
magnitude and direction into account
a) Use the component form of conservation of
momentum:
Px,initial = Px,final
(X.F.5,6)
Py,initial = Py,final
Note: True for elastic or inelastic collisions
x
“Odd – Job”
Example Problem
y
Conservation of Momentum:


Pbefore  Pafter
vOJ
ice
BEFORE
vhat
36.90x
ice
AFTER
x–direction: 0 = phat,x + pOJ,x
y–direction: 0 = phat,y + pOJ,y
We are interested in the horizontal recoil velocity of the bad
guy, so from the x – component equation we have:
 pOJ,x = – phat,x  mOJvOJ = – mhatvhat,x = – mhatvhat cos36.90
 vOJ = – mhatvhat cos36.90 / mOJ
= – (4.50 kg)(22.0 m/s)cos36.90 / 120 kg = – 0.66 m/s
Example Problem #8.20
vA = 0
A
vB = 0
B
vA’
vB’
A
B
BEFORE
AFTER
(a) Conservation of momentum: Ptotal,before = Ptotal,after
mAvA + mBvB = mAvA’ + mBvB’
0 = mAvA’ + mBvB’  mAvA’ = –mBvB’
vA’ = –mBvB’ / mA = –[(3.00 kg)(1.20 m/s)] / 1.00 kg,
= –3.6 m/s
(b) Conservation of mechanical energy:
Kbefore + Uel,before = Kafter + Uel,after
0 2
0 2
1/2mAvA + 1/2mBvB + Uel,before
0
2
2
= 1/2mAvA’ + 1/2mBvB’ + Uel,after
Uel,before
= 1/2(1 kg)(–3.6 m/s)2 + 1/2(3 kg)(1.2m/s)2
= 8.6 J
Example Problem #8.50
(a) xcm = (m1x1 + m2x2) / (m1 + m2)
= (1800 kg)(40.0 m) / 3000 kg
= 24.0 m ahead of m1
(or 16.0 m behind m2)
x=0
(b) P = m1v1 + m2v2
= (1200 kg)(12.0 m/s)
+ (1800 kg)(20.0 m/s)
= 5.04  104 kgm/s
(c) vcm = (m1v1 + m2v2) / (m1 + m2)
= [(1200 kg)(12.0 m/s) + (1800 kg)(20.0 m/s)/3000 kg
= 16.8 m/s
Example Problem
Consider a solid disk rolling without slipping down an inclined
plane. What is its velocity at the bottom of the plane if it starts
from rest (~no friction).
Ei = mgh.
Ef = 1/2mv2 + (1/2)Iw2.
But:
h
I = 1/2mR2, and
w  v/R
So:
mgh = 1/2mv2 + (1/2)(1/2mR2v/R)2 = (3/4)mv2, or
v = SQRT(4/3gh).
Example Problem
A uniform solid ball of mass m and radius R rolls without
slipping down a plane inclined at an angle q. Using
dynamics (Newton’s 2nd Law), find the final speed.
vcm
h
q
E1 = mgh = mgsinq
E2 = 1/2mv2 + 1/2Iw2 = 1/2mv2 + 1/2(2/5mR2)(v/R)2
= 1/2mv2( 1+ 2/5) = 7/10mv2 = mgsinq;
v = 1.2(gsinq)1/2